phys121_s12_hw1_sol_mp_1

# phys121_s12_hw1_sol_mp_1 - Homework 1 PHYS 121 2.5 2.9 2.10...

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1 (Due Date: 02/06/2012) Homework # 1, PHYS 121 Spring 2012 Mastering Physics: P2.17, P2.22, P2.19, P2.25, P2.37, P2.67, P2.72 P2.17. Prepare: The graph in Figure P2.17 shows distinct slopes in the time intervals: 0 – 1s, 1 s – 2 s, and 2 s – 4 s. We can thus obtain the velocity values from this graph using v = x / t . Solve: (a) (b) There is only one turning point. At t = 2 s the velocity changes from +20 m/s to 10 m/s, thus reversing the direction of motion. At t = 1 s, there is an abrupt change in motion from rest to +20 m/s, but there is no reversal in motion. Assess: As shown above in (a) , a positive slope must give a positive velocity and a negative slope must yield a negative velocity. P2.19. Prepare: Please refer to Figure P2.19. Since displacement is equal to the area under the velocity graph between t i and t f , we can find the car’s final position from its initial position and the area. Solve: (a) Using the equation x f = x i + area of the velocity graph between t i and t f , 2 s 3 s 4 s 3 s 10 m area of trapezoid between 0 s and 2 s 1 10 m (12 m/s 4 m/s)(2 s) 26 m 2 10 m area of triangle between 0 s and 3 s 1 10 m (12 m/s)(3 s) 28 m 2 area between 3 s and 4 s 1 28 m ( 4 m/s 2 x x xx  )(1 s) 26 m (b) The car reverses direction at t = 3 s, because its velocity becomes negative. Assess: The car starts at x i = 10 m at t i = 0. Its velocity decreases as time increases, is zero at t = 3 s, and then becomes negative. The slope of the velocity-versus-time graph is negative which means the car’s acceleration is negative and a

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phys121_s12_hw1_sol_mp_1 - Homework 1 PHYS 121 2.5 2.9 2.10...

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